Pagini recente »
Monitorul de evaluare
|
tema_10_2024_1
|
Atașamentele paginii Clasament 2015-05-12-clasa-5-tema-38
|
OJI 2023 Clasa a VI-a - Antrenament - FFA v2.1
|
Cod sursă (job #546599)
Cod sursă (job
#546599)
#include <fstream>
#include<bits/stdc++.h>
using namespace std;
ifstream fin ("fotografie.in");
ofstream fout ("fotografie.out");
int aa[1002], mx=1, my=1, mz=1, p, q, m, n, K, sm, val;
char a[1002][1002], b[1002][1002];
void rezolva(int x, int y){
for (int i = x; i > x-p; --i)
for (int j = y; j > y-q; --j)
if (a[i][j] != b[i-(x-p+1)][j-(y-q+1)])
return ;
fout<<x-p+1<<' '<<y-q+1<<'\n';
}
int main()
{
fin >> n >> m;
K=46103;
for (int i = 0; i<n; ++i)fin>>a[i];
fin >>p>>q;
for (int i = 0; i < p; ++ i)fin >> b[i];
for (int j = 0; j < q; ++ j)
for (int i = 0; i < p; ++ i) sm = (sm * 26 + (b[i][j] - 'a')) % K;
for (int i = 1; i < p; ++ i)mx = (mx * 26) % K;
mz = (mx * 26) % K;
for (int i = 1; i < q; ++ i)my = (my * mz) % K;
for (int j = 0; j < m; ++ j)
for (int i = 0; i < p - 1; ++ i)
aa[j] = (aa[j] * 26 + (a[i][j] - 'a')) % K;
for (int i = p - 1; i < n; ++ i)
{
for (int j = 0; j < m; ++ j)
aa[j] = (aa[j] * 26 + (a[i][j] - 'a')) % K;
val = 0;
for (int j = 0; j < q - 1; ++ j)
val = (val * mz + aa[j]) % K;
for (int j = q - 1; j < m; ++ j)
{
val = (val * mz + aa[j]) % K;
if (val == sm)rezolva(i, j);
val = ((val - my * aa[j - q + 1]) % K);
if (val < 0)val += K;
}
for (int j = 0; j < m; ++ j)
{
aa[j] = (aa[j] - mx * (a[i - p + 1][j] - 'a')) % K;
if (aa[j] < K)aa[j] += K;
}
}
return 0;
}